Answer:
The correct options are: Interquartile ranges are not significantly impacted by outliers. Lower and upper quartiles are needed to find the interquartile range. The data values should be listed in order before trying to find the interquartile range. The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range. The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median
P(t)=6t
A(p)=πp^2, since p(t)=6t
A(t)=π(p(t))^2
A(t)=π(6t)^2
A(t)=36πt^2, so when t=8 and approximating π≈3.14
A(8)≈36(3.14)(8^2)
A(8)≈36(3.14)64
A(8)≈7234.56 u^2
The pair of points defines a line with a slope of -1 is C. (10,3) and (15,-2)
Answer:
(6,7) is a solution
Step-by-step explanation:
To determine if (6,7) is a solution, we substitute the point in and see if the inequality is true
15x+11y>12
15(6) + 11(7) >12
90+77 > 12
167>12
This is true so (6,7) is a solution
